Fermi-Dirac statistics — fermions, Fermi energy: Worked examples

This page is a drill. We take the parent topic and hit every kind of question it can throw at you: every regime of energy relative to the chemical potential, the two temperature limits, degenerate inputs, real-world numbers, and an exam-style twist. First we lay out the full grid of cases, then we work one example per cell.

Before anything, one reminder of the two objects we keep using:

The scenario matrix

Every cell below is a distinct thing the topic can test. Each numbered example targets one (or two) cells.

Cell	Case class	What makes it different	Example
A	$ε = μ$ exactly	exponent is $0$ ; answer is $T$ -independent	Ex 1
B	$ε > μ$ (above the sea)	positive exponent; $f < \frac{1}{2}$ , decays	Ex 2
C	$ε < μ$ (below the sea)	negative exponent; $f > \frac{1}{2}$	Ex 2
D	$T \to 0$ limit	step function; degenerate answer	Ex 3
A∩D	$ε = μ$ and $T \to 0$ (the corner)	conventional $f = \frac{1}{2}$ by limit	Ex 1 (note)
E	High-energy tail $ε - μ ≫ k_{B} T$	reduces to Maxwell-Boltzmann	Ex 4
F	Symmetry of $f$ about $μ$ (particle↔hole)	$f (μ + δ) + f (μ - δ) = 1$	Ex 5
G	Real-world density → $E_{F}$ , $T_{F}$ , $v_{F}$	plug real $n$ (sodium)	Ex 6
H	Degenerate/zero input: $n \to 0$ or $n$ doubled	scaling law $E_{F} \propto n^{2/3}$	Ex 7
I	Inverting $f$ to find $T$ (exam twist)	solve for the unknown temperature	Ex 8
J	Width of thermal smear $\approx 4.4 k_{B} T$	geometric read of the curve	Ex 9 (figure)

Example 1 — Cell A: right on the Fermi level

Worked example Occupation exactly at

$ε = μ$ A quantum state sits at energy exactly equal to the chemical potential, at $T = 300$ K. What fraction of the time is it occupied? Does the answer change if $T = 3000$ K?

Forecast: Guess before reading — will hotter mean fuller, emptier, or no change?

Step 1. Write the exponent $x = (ε - μ) / k_{B} T$ . Since $ε = μ$ , the numerator is $0$ , so $x = 0$ for any $T > 0$ . Why this step? The whole temperature dependence lives inside that exponent. If the numerator is zero, $T$ cannot matter.

Step 2. Plug in: $f = \frac{1}{e ^{0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$ . Why this step? $e^{0} = 1$ exactly; this is the definitional anchor of $μ$ .

Verify: $\frac{1}{2}$ is dimensionless and lies in $[0, 1]$ ✓. It did not change with $T$ — so the correct forecast was "no change". This is precisely how experimenters locate $μ$ : find the energy that stays half-filled.

Example 2 — Cells B & C: above and below the sea

Worked example Two states straddling

$μ$ At $T = 300$ K take two states: one above $μ$ by $0.05$ eV, one below $μ$ by $0.05$ eV. Find $f$ for each.

Forecast: One will be more than $\frac{1}{2}$ , one less. Which is which, and are they related?

Step 0 (the thermal scale). From the box above, at $T = 300$ K, $k_{B} T = (8.617 \times 1 0^{- 5}) (300) = 0.02585$ eV. Why this step? We will divide the eV offset by this eV energy, so both must be in eV — that is exactly what the eV/K face of $k_{B}$ buys us.

Step 1 (Cell B, above). $x_{+} = \frac{+ 0.05}{0.02585} = 1.934$ . Then $f_{+} = \frac{1}{e ^{1.934} + 1} = \frac{1}{6.92 + 1} = 0.1263$ . Why this step? Above $μ$ the numerator is positive, so $x > 0$ , so $e^{x} > 1$ , so $f < \frac{1}{2}$ : states above the sea are mostly empty.

Step 2 (Cell C, below). $x_{-} = \frac{- 0.05}{0.02585} = - 1.934$ . Then $f_{-} = \frac{1}{e ^{- 1.934} + 1} = \frac{1}{0.1445 + 1} = 0.8737$ . Why this step? Below $μ$ the numerator is negative, so $e^{x} < 1$ , so $f > \frac{1}{2}$ : states below the sea are mostly full.

Verify: $f_{+} + f_{-} = 0.1263 + 0.8737 = 1.000$ exactly. This is the particle–hole symmetry (Cell F) sneaking in: an empty slot a distance $δ$ above $μ$ is as likely as a filled slot $δ$ below. Both values sit in $[0, 1]$ ✓.

Example 3 — Cell D: the $T = 0$ step

Worked example Cold metal, sharp edge

At $T = 0$ , evaluate $f$ for a state $0.5$ eV below $E_{F}$ and a state $0.5$ eV above $E_{F}$ .

Forecast: At absolute zero the fuzzy edge should vanish — guess the two numbers.

Step 1 (below). As $T \to 0$ , for $ε < E_{F}$ the exponent $(ε - E_{F}) / k_{B} T \to \frac{negative}{0 ^{+}} \to - \infty$ , so $e^{- \infty} = 0$ and $f = \frac{1}{0 + 1} = 1$ . Why this step? Below the Fermi energy every state is filled — the exclusion principle forces electrons to stack up rather than pile into the ground state.

Step 2 (above). For $ε > E_{F}$ the exponent $\to + \infty$ , so $e^{+ \infty} = \infty$ and $f = \frac{1}{\infty + 1} = 0$ . Why this step? Above the sea there is nothing to fill those slots at zero temperature.

Verify: We recovered the step function: $f = 1$ below, $f = 0$ above, with the jump at $E_{F}$ . Consistent with Ex 1's half-filling limit approached from either side ✓ (and exactly at $E_{F}$ the value is $\frac{1}{2}$ , the A∩D corner).

Example 4 — Cell E: the classical (Maxwell–Boltzmann) tail

Worked example When Fermi-Dirac becomes classical

A state lies $0.30$ eV above $μ$ at $T = 300$ K. Show $f$ is well-approximated by the Maxwell-Boltzmann form $e^{- (ε - μ) / k_{B} T}$ , and find the relative error.

Forecast: Far up the tail, does the " $+ 1$ " still matter?

Step 1 — algebraic reduction. Multiply numerator and denominator of $f$ by $e^{- x}$ (with $x = (ε - μ) / k_{B} T$ ): $f = \frac{1}{e ^{x} + 1} = \frac{e ^{- x}}{1 + e ^{- x}} .$ Now when $x ≫ 1$ , the term $e^{- x}$ in the denominator is tiny, so $1 + e^{- x} \to 1$ and hence $f ≃ \frac{e ^{- x}}{1} = e^{- x} = e^{- (ε - μ) / k_{B} T} .$ Why this step? This is the exact algebraic bridge from Fermi-Dirac to Maxwell-Boltzmann — the approximation is "drop $e^{- x}$ against $1$ in the denominator", which is legitimate precisely because $e^{- x}$ is small in the high-energy tail.

Step 2 — numbers. With $k_{B} T = 0.02585$ eV (same $T = 300$ K as before, eV/K face of $k_{B}$ ): $x = \frac{0.30}{0.02585} = 11.60$ . Then $f_{MB} = e^{- x} = e^{- 11.60} = 9.14 \times 1 0^{- 6}$ , while the exact $f = \frac{1}{e ^{11.60} + 1} = 9.14 \times 1 0^{- 6}$ (agree to five figures). Why this step? This is why classical statistics works for dilute/high-energy states — it is the $ε - μ ≫ k_{B} T$ limit of Fermi-Dirac.

Verify: Relative error $= \frac{f _{MB} - f}{f} = e^{- x} \approx 9 \times 1 0^{- 6}$ , i.e. about $0.0009%$ — utterly negligible ✓.

Example 5 — Cell F: particle–hole symmetry, made general

Example 6 — Cell G: real metal (sodium)

Worked example Fermi energy, temperature, velocity of sodium

Sodium has $n = 2.65 \times 1 0^{28} m^{- 3}$ free electrons. Find $E_{F}$ (in eV), $T_{F}$ , and $v_{F}$ .

Forecast: Copper (parent note) gave $\sim 7$ eV at $n = 8.5 \times 1 0^{28}$ . Sodium is roughly $3 \times$ less dense — guess whether $E_{F}$ is bigger or smaller.

Step 1 — build $k_{F}$ . $k_{F} = (3 π^{2} n)^{1/3} = (3 \cdot 9.8696 \cdot 2.65 \times 1 0^{28})^{1/3} = (7.846 \times 1 0^{29})^{1/3} = 9.225 \times 1 0^{9} m^{- 1}$ . Why this step? $E_{F}$ is cleanest through the Fermi wavevector $k_{F}$ ; the cube root undoes the sphere volume $\frac{4}{3} π k_{F}^{3}$ . Keep full precision in $k_{F}$ — squaring it later magnifies any rounding.

Step 2 — energy in joules. $E_{F} = \frac{ℏ ^{2} k _{F}^{2}}{2 m} = \frac{( 1.055 \times 1 0 ^{- 34} ) ^{2} ( 9.225 \times 1 0 ^{9} ) ^{2}}{2 ( 9.11 \times 1 0 ^{- 31} )} = 5.20 \times 1 0^{- 19} J$ . Why this step? $ε = ℏ^{2} k^{2} /2 m$ is the free-particle energy; evaluating at $k = k_{F}$ gives the top of the sea. (Rounding $k_{F}$ to only 3 figures shifts this by a few $%$ , so we carried the fuller value.)

Step 3 — convert to eV. Divide by the conversion factor $1 eV = 1.602 \times 1 0^{- 19} J$ : $E_{F} = \frac{5.20 \times 1 0 ^{- 19} J}{1.602 \times 1 0 ^{- 19} J/eV} = 3.24 eV .$ Why this step? Joules are the SI energy; eV is the natural scale for electrons — the conversion factor is what makes the number human-readable.

Step 4 — temperature & velocity. $T_{F} = E_{F} / k_{B} = \frac{5.20 \times 1 0 ^{- 19} J}{1.381 \times 1 0 ^{- 23} J/K} = 3.77 \times 1 0^{4} K$ . $v_{F} = ℏ k_{F} / m = \frac{( 1.055 \times 1 0 ^{- 34} ) ( 9.225 \times 1 0 ^{9} )}{9.11 \times 1 0 ^{- 31}} = 1.068 \times 1 0^{6} m/s$ . Why this step? $T_{F}$ tells us how degenerate the gas is ( $T_{F} ≫ 300$ K ⇒ deeply degenerate); $v_{F}$ is the speed of electrons at the sea surface. Note we use $E_{F}$ in joules here so it cancels the joules in $k_{B}$ .

Verify: Units: $J / (J/K) = K$ ✓; $v_{F}$ is $\sim 0.4%$ of light speed (non-relativistic, so free-electron formula is valid) ✓. Lower density than copper ⇒ lower $E_{F}$ (3.24 < 7 eV), matching the $E_{F} \propto n^{2/3}$ scaling — correct forecast. (Standard tables quote sodium $E_{F} \approx 3.24$ eV.)

Example 7 — Cell H: degenerate inputs (double the density, empty the box)

Example 8 — Cell I: exam twist, solve for temperature

Worked example Given the occupation, find

$T$ A state $0.10$ eV above $μ$ is observed to have occupation $f = 0.20$ . What temperature is the system at?

Forecast: More thermal smearing (higher $T$ ) pushes $f$ at a fixed offset closer to $\frac{1}{2}$ . Since $0.20 < \frac{1}{2}$ , guess: is $T$ small or large?

Step 1 — invert $f$ . From $f = \frac{1}{e ^{x} + 1}$ : $e^{x} = \frac{1}{f} - 1 = \frac{1}{0.20} - 1 = 4$ , so $x = ln 4 = 1.386$ . Why this step? We know the output $f$ and want the input $x$ ; algebraically isolating $e^{x}$ then taking $ln$ is the inverse of the distribution.

Step 2 — solve for $T$ . $x = \frac{ε - μ}{k _{B} T} \Rightarrow T = \frac{ε - μ}{k _{B} x} = \frac{0.10 eV}{( 8.617 \times 1 0 ^{- 5} eV/K ) ( 1.386 )} = 837 K$ . Why this step? We use $k_{B}$ in eV/K (the face we built at the top, $8.617 \times 1 0^{- 5}$ ) so the eV in the numerator cancels, leaving kelvin.

Verify: Plug back: $x = (0.10) / (8.617 \times 1 0^{- 5} \cdot 837) = 1.386$ , and $f = 1/ (e^{1.386} + 1) = 1/ (4 + 1) = 0.20$ ✓. This $T$ is well below $T_{F}$ , consistent with a mostly-empty state above $μ$ .

Example 9 — Cell J: geometry of the thermal smear (figure)

Worked example How wide is the fuzzy edge?

At $T = 1000$ K, over what energy width does $f$ fall from $0.90$ to $0.10$ ? Read it off the curve.

Forecast: The parent note claims $\approx 4.4 k_{B} T$ . Guess the width in eV first.

How to read the figure: the thick blue S-shaped curve is $f$ plotted against the scaled energy $x = (ε - μ) / k_{B} T$ . The vertical dashed gray line marks $ε = μ$ ( $x = 0$ ), where the gray dot sits at the half-filled value $f = 0.5$ . The two round orange markers on the curve's shoulders sit at $f = 0.90$ (upper-left) and $f = 0.10$ (lower-right); these are the two occupation levels the question asks about. The green horizontal double-headed arrow with its shaded band spans the horizontal distance between those two markers — that span is the width we want, measured in units of $k_{B} T$ .

Step 1 — the two edge exponents. $f = 0.90 \Rightarrow e^{x} = \frac{1}{0.9} - 1 = 0.111 \Rightarrow x = ln (0.111) = - 2.197$ . $f = 0.10 \Rightarrow e^{x} = 9 \Rightarrow x = + 2.197$ . Why this step? These $x$ values are exactly the horizontal positions of the two orange shoulder markers in the figure. Their span in $x$ is $2.197 - (- 2.197) = 4.394$ — the length of the green band.

Step 2 — convert to energy. The width in scaled units is $Δ x = 4.394$ , so $Δ ε = 4.394 k_{B} T$ . With $k_{B} T = (8.617 \times 1 0^{- 5} eV/K) (1000 K) = 0.0862$ eV: $Δ ε = 4.394 \times 0.0862 = 0.379$ eV. Why this step? $x$ is measured in units of $k_{B} T$ , so multiplying the $x$ -span (the band's length) by $k_{B} T$ converts the "shoulder width" from dimensionless to eV. This is the quantitative read-off the figure was set up to deliver.

Verify: $4.394 \approx 4.4$ , matching the parent note's $4.4 k_{B} T$ claim ✓. Numerically $Δ ε = 0.379$ eV at $1000$ K. In a metal with $E_{F} \sim 3$ – $7$ eV this $0.38$ eV smear is a thin sliver near the top — only electrons in that band are thermally/electrically active. Units: (dimensionless)×eV = eV ✓, closing cell J.

Recall Quick self-test

A state at $ε - μ = - k_{B} T$ has what occupation? ::: $f = 1/ (e^{- 1} + 1) = 1/1.368 = 0.731$ If $f = 0.5$ and $ε > 0$ , where is $μ$ ? ::: exactly at $ε$ (Cell A) Doubling $n$ multiplies $E_{F}$ by? ::: $2^{2/3} \approx 1.587$ (Cell H) The high-energy tail of $f$ reduces to which classical law? ::: Maxwell-Boltzmann $e^{- (ε - μ) / k_{B} T}$ (Cell E)

See also: Density of States (turns $f$ into particle numbers), Sommerfeld Expansion (how $μ$ drifts from $E_{F}$ at finite $T$ ), and Electron Degeneracy Pressure & White Dwarfs (what the filled sea holds up).